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Moment Inertia of a cube

  1. Jan 12, 2010 #1
    To which of the two cubes has a larger moment of inertia?
    untitled.JPG
    I think it's the right one, is it correct?
    How can I explain that without using the parallel axis theorem?
     
  2. jcsd
  3. Jan 12, 2010 #2

    berkeman

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    Staff: Mentor

    Why do you say the right one? Are you familiar with the relevant equation for calculating the moment of inertia?
     
  4. Jan 12, 2010 #3
    what relevant equation?

    I think it's the right one because We know that the minimal moment of inertia is throw the principal axes that goes throw the center of mass. in the right one , the rotation isn't throw the principal axes . there is also the following theorem :

    The moment of inertia about an arbitrary axis is equal to the
    moment of inertia about a parallel axis passing through the
    center of mass plus the moment of inertia of the body about
    the arbitrary axis, taken as if all of the mass M of the body
    were at the center of mass.

    Am I wrong?
     
    Last edited: Jan 12, 2010
  5. Jan 12, 2010 #4

    berkeman

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    There may be a shortcut way to tell which has a higher moment of inertia, but for me, I'd need to calculate it. I'd use the standard definition of the Mmoment of inertia, and evaluate thge integral for the diagonal case. I don't think you can use the parallel axis theorm, since the two axes are not parallel.

    I'd do the 2-D case first, to see if it offered some intuition. That is, the moment of inertia for a flat rectangular sheet, with the axes going straight versus diagonal.
     
  6. Jan 12, 2010 #5

    vela

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    I think they'll turn out to be equal. Try computing the moment of inertia tensor.
     
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